## BdMO National Higher Secondary 2009/5

Moon
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BdMO National Higher Secondary 2009/5
Problem 5:
In triangle $ABC,\ \angle A = 90\circ$. $M$ is the midpoint of $BC$. Choose $D$ on $AC$ such that $AD=AM$. The circumcircles of triangles $AMC$ and $BDC$ intersect at $C$ and at a point $P$. What is the ratio: $\frac {\angle ACB}{\angle PCB}=?$
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

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Ashfaq Uday
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Joined:Tue Sep 27, 2011 12:18 am

### Re: BdMO National Higher Secondary 2009/5

I think the answer is 2. P is the incenter of ABC. Can somebody please post the proof?? I am missing the easier solution and keeping messing around with many things. Proved a lot of unnecessary contents.

Labib
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Joined:Thu Dec 09, 2010 10:58 pm

### Re: BdMO National Higher Secondary 2009/5

Would love some hints for this one!!
Please Install $L^AT_EX$ fonts in your PC for better looking equations,
Learn how to write equations, and don't forget to read Forum Guide and Rules.

"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes

sourav das
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### Re: BdMO National Higher Secondary 2009/5

Try to work backward and see if you can find any congruent triangles.....
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

Ashfaq Uday
Posts:21
Joined:Tue Sep 27, 2011 12:18 am

### Re: BdMO National Higher Secondary 2009/5

let$O_{1}=$circumcenter of $\triangle AMC$$X=O_{1}P\cap AB,Y=BA\cap CO_{1}$$O_{1}AXM$is a rhombus. $\Rightarrow AE=EM\Rightarrow AP=PM\Rightarrow \angle PYA=\angle PCA=\angle PAM=\angle PCM$

sourav das
Posts:461
Joined:Wed Dec 15, 2010 10:05 am
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### Re: BdMO National Higher Secondary 2009/5

@Ashfaq Uday
My questions

i) Reason of claiming $O_1AXM$ a rohmbus.
ii) Define point $E$.

Please make your solution much more detailed so that everyone can understand and learn the way of your solution. It helps both the reader and solver.

I'm giving my solution:
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )